Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2715322

Let CD = AB = 1 unit. a = 180-30-40 a = 110° a is angle CAD. (1/sin110) = (b/sin30) b = 0.5320888862 units. b is AC. Therefore, the required angle, x is; (0.5320888862/sinx) = (1/sin40) x = 20°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2715303

Sir Mike Ambrose is the author of the question. The equation of the curve is; y = -½(x+2)²(x-4) y = -½(x²+4x+4)(x-4) y = -½(x³-12x-16) y = -½(x³)+6x+8 It implies; dy/dx = -½(3x²)+6 For dy/dx = 0 x = 2 Therefore; y = -½(2³)+6*2+8 y = 16 At the maximum turning point P. P(2, 16) Shaded area is; Area under the curve at point (0, 0) and (4, 0) - Area triangle with height (64/√(65)) units and base (8/√(65)) units. = 48 - ½((64/√(65))(8/√(65))) = 48 - ½(64*8)/65...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2714408

Pink area is; ½(area square with side 4 cm - area quarter circle with radius 4 cm) = ½(4*4 - ¼(16π)) = ½(16 - 4π) = 8 - 2π = 2(4-π) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2713706

Let the length (base) of the ascribed rectangle be 4 units. Therefore its width/breadth/height is 2 units. tana = 4/2 a = atan(2)° b = atan(½)° sin(atan(½)) = c/4 c = 3.577708764 units. d² = 4²+2² d = 4.472135955 units. e = d-c e = 0.894427191 units. f = 180-2atan(½) f = 126.8698976458° g² = 2-2cos126.8698976458 g = 1.788854382 units. h = g-e h = 0.894427191 units. j = 90-atan(2) j = 26.5650511771° k = 2j k = 53.1301023542° cos53.1301023542 = l/1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2713691

Let the radius of the ascribed quarter circle be 1 unit. It implies; R = 1 unit. 2a² = 1² a² = ½ a = ½√(2) units. a is the side length of the inscribed square. b = 2R b = 2 units. b is the diameter of the complete circle. c² = 2²+(½√(2))²-2*2*½√(2)cos45 c² = 4+½-2 c = √(5/2) c = ½√(10) units. c = 1.5811388301 units. (1.5811388301/sin45) = (2/sind) d = 63.4349488229° sin63.4349488229 = e/(0.5√(2)) e = 0.632455532 units. cos63.4349488229 = f/(0.5√(2)) f = 0.316...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2713186

Sir Mike Ambrose is the author of the question. The vertices of area purple is; (6, 12) ((24/5), (36/5)) ((180/29), (168/29)) (12, (21/2)) Area purple is; ½((6, 12) ((24/5), (36/5)) ((180/29), (168/29)) (12, (21/2)) (6, 12)) = ½((-72/5)+(-2448/145)+(-126/29)+81) = ½((-2088-2448-630+11745)/145) = ½(6579/145) = 6579/290 square unit.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2712637

Notice. MN = CQ = 6 units. MQ = 6+2 MQ = 8 units. a² = 8²+6² a = √(100) a = 10 units. a is CM. 3 - 6 r - 10 Cross Multiply. 6r = 30 r = 5 units. r is the radius of the inscribed circle. 5² = 3²+b² b = √(25-9) b = √(16) b = 4 units. c = r+b c = 9 units. c is BQ. d = c+6 d = 15 units. d is BC. tane = 6/8 e = atan(3/4)° e is angle CMQ = angle BAC. tan(atan(3/4)) = 15/f ¾ = 15/f 3f = 4*15 f = 20 units. f is AB. g² = 20²+15² g = √(400+225)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2711951

Calculating x. Let the base of the ascribed triangle be 1 unit. a = 180-15-45 a = 120° (1/sin120) = (b/sin15) b = 0.2988584907 units. c = 180-15-30 c = 135° (0.2988584907/sin15) = (d/sin135) d = 0.8164965809 units. e = ½(d) e = 0.4082482905 units. f² = 0.4082482905²+0.2988584907²-2*0.4082482905*0.2988584907cos30 f = 0.2113248654 units. (0.2113248654/sin30) = (0.2988584907/sing) g = 45° Therefore, angle x is; x = 180-30-45-45 x = 180-120 x = 60°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2711850

Notice. Calculating the required angle, alpha. 4 units is the radius of the quarter circle. It implies; 3² = 4²+4²-2*4*4cosa 9 = 32-32cosa 32cosa = 32-9 a = acos(23/32) a = 44.0486256741° b = ½(180-a) b = 67.975687163° c = 90-b c = 22.024312837° d² = 4²+2²-2*4*2cos22.024312837 d = 2.273236245 units. It implies, the required angle, alpha is; Let it be f. (2.273236245/sin22.024312837) = (4/sine) e = 41.2886037977 f = 180-e f = 138.7113962023° Again, f...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2711833

Calculating Area Shaded. Let r be the radius of the semi circle. Calculating r. ((6r/√(10))-2)²=(r/2)²+(3r/2)² 72r²-48√(10)r+80=50r² 22r²-48√(10)r+80= 0 11r²-2⁴√(10)r+40= 0 Therefore; r ≠ 0.57496 cm. r = 6.32455 cm. The height, h of the shaded triangle is; h = r/2 cm. The base of the shaded triangle is; b = 2r cm. And r = 6.32455 cm. Therefore; h = 6.32455/2 h = 3.162275 cm. And b = 2(6.32455) b = 12.6491 cm. It implies; Area shaded triangle...

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